# Full CI computing power

1. Aug 12, 2013

For a molecule of $n$ nuclei and $x$ total electrons, how can I work out how much computing power or processing time is required for an exact Full CI calculation for that molecule?

2. Aug 13, 2013

### DrDu

For an exact full CI you need a complete MO basis which contains an infinity of molecular orbitals and you can form an infinity of Slater determinants from it. Hence you will need infinite computing power for more than one electron in general.
So you have to be less ambitious. E.g. you can chose a certain finite basis set, e.g. VQZ. Let's say you have 100 basis functions per atom (quite a big basis already) (this will also depend on the type of atom). If you have 10 atoms with 10 electrons each, you can form 100 out of 1000 Slater determinants (roughly 10^300) so this will still take almost forever.

3. Aug 13, 2013

Hang on, please clarify. Ok so as you've pointed out an exact solution will take infinite time, i.e. impossible, except for one-electron systems. If you have a certain basis set with k basis functions per nucleus, and n nuclei with x total electrons ... I still don't get how you are proposing to work out the amount of computing power needed? Thanks for the guidance. Can you do an example, maybe, let there be k basis functions, but now we're talking about an H2O molecule (n=3, x=10), how much processing power/time is needed?

4. Aug 14, 2013

### DrDu

So if you have k basis functions and x electrons, you have 2k spin orbitals. As each Slater determinant has to be made up from x different spin orbitals, you can form $$m=\begin{pmatrix} 2k\\x \end{pmatrix}\approx x^{2k}$$ different Slater determinants (you could classify them by spin and symmetry to reduce the number, but we are interested in the gross picture). So basically your Hamiltonian becomes an $$m \times m$$ matrix which you have to diagonalize. The matrix will be sparse as the two electron matrix elements vanish between most of the Slater determinants but still the size increases factorially with the size of the system.

5. Aug 14, 2013